A) \[a<2\]
B) \[2\le a\le 3\]
C) \[3<a\le 4\]
D) \[a>4\]
Correct Answer: A
Solution :
Given equation is \[{{x}^{2}}-2ax+{{a}^{2}}+a-3=0\] If roots are real, then \[D\ge 0\] Þ \[4{{a}^{2}}-4({{a}^{2}}+a-3)\ge 0\,\,\,\Rightarrow \,\,-a+3\ge 0\] Þ \[a-3\le 0\,\,\,\Rightarrow \,\,\,a\le 3\] As roots are less than 3, hence \[f(3)>0\] \[9-6a+{{a}^{2}}+a-3>0\,\,\,\Rightarrow {{a}^{2}}-5a+6>0\] \[\Rightarrow (a-2)(a-3)>0\Rightarrow \] either\[a<2\] or \[a>3\] Hence \[a<2\] satisfy all.
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